Can you prove or disprove that the sequence $\{\sin (nx)\}$ has a pointwise almost everywhere convergent subsequence with respect to the Lebesgue measure on $\mathbb{R}$ ?
Edit:
I am adding my thoughts here as a motivation, because otherwise this question will hit the bottom. So initially I thought whether $\{\sin{nx}\}$ is convergent in $L_1([0,\pi])$ or on other finite interval in $\mathbb R$. Then I computed the integral $\|\sin{nx}\|_1=\int\limits_{0}^{\pi}{|\sin{nx}|dx}=n\int\limits_{0}^{\pi/n}{\sin{nx}dx}=2$. Therefore $\|\sin{nx}\|_1\rightarrow 2$ but this doesn't tell me to which function converges. Then I thought about convergence in $L_2([0,\pi])$, but obviously it is not convergent there. This is because it is part of the orthonormal basis in $L_2$ (up to a constant), so it is not even Cauchy sequence. But still it is weakly convergent in $L_2$, because from Bessel's inequality it follows that $\langle f,\sin{nx}\rangle\rightarrow 0$ for each $f\in L_2([0,\pi])$. Finally, I decided to check whether $\{\sin{nx}\}$ has a convergent a.e subsequence. If there is no such sequence, then it can not be convergent in $L_1$. And if there is such subsequence, then by Lebesgue DCT it will follow that it converges in $L_1$ (and probably it might be the limit of the whole sequence). I just didn't see that LDCT will work again for $L_2$ : if there is a convergent a.e subsequence, then it should converge in $L_2$ also, but this is impossible since $\{\sin{nx}\}$ is orthonormal (up tp a constant) and no subsequence of $\{\sin{nx}\}$ is Cauchy in $L_2$. This is actually the answer of @Julián Aguirre.
Now I have another Question: Is it possible to prove that there is no convergent pointwise a.e subsequence of $\{\sin{nx}\}$ using only first year calculus and knowing of course what is a set with Lebesgue measure $0$ in $\mathbb R$ ?
What is obvious is that for each $x$ there is a convergent subsequence since $\{\sin{nx}\}$ is bounded. But all $x\in [0,\pi]$ are uncountable set, so we can not apply for example Cantor diagonal argument.
Best Answer
A different proof. If $\{\sin(n_k\,x)\}$ converges a.e., then $$ (\sin(n_{k+1}\,x)-\sin(n_k\,x))^2 $$ converges to $0$ a.e. By the dominated convergence theorem $$ \lim_{k\to\infty}\int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=0, $$ but $$ \int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=\pi. $$